c 2015 Society for Industrial and Applied Mathematics

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1 SIAM J. DISCRETE MATH. Vol. 29, No. 2, pp c 2015 Society for Industrial and Applied Mathematics TSP TOURS IN CUBIC GRAPHS: BEYOND 4/3 JOSÉ CORREA, OMAR LARRÉ, AND JOSÉ A. SOTO Abstract. After a sequence of improvements Boyd et al. [TSP on cubic and subcubic graphs, Integer Programming and Combinatorial Optimization, Lecture Notes in Comput. Sci. 6655, Springer, Heidelberg, 2011, pp ] proved that any 2-connected graph whose n vertices have degree 3, i.e., a cubic 2-connected graph, has a Hamiltonian tour of length at most (4/3)n, establishing in particular that the integrality gap of the subtour LP is at most 4/3 for cubic 2-connected graphs and matching the conjectured value of the famous 4/3 conjecture. In this paper we improve upon this result by designing an algorithm that finds a tour of length (4/3 1/61236)n, implying that cubic 2-connected graphs are among the few interesting classes of graphs for which the integrality gap of the subtour LP is strictly less than 4/3. With the previous result, and by considering an even smaller ɛ, weshow that the integrality gap of the TSP relaxation is at most 4/3 ɛ even if the graph is not 2-connected (i.e., for cubic connected graphs), implying that the approximability threshold of the TSP in cubic graphs is strictly below 4/3. Finally, using similar techniques we show, as an additional result, that every Barnette graph admits a tour of length at most (4/3 1/18)n. Key words. traveling salesman problem, cubic graphs, integrality gap AMS subject classifications. 05C85, 05C45, 68W25, 68W40 DOI / Introduction. The traveling salesman problem (TSP) in metric graphs is a landmark problem in combinatorial optimization and theoretical computer science. Given a graph in which edge-distances form a metric the goal is to find a tour of minimum length visiting each vertex exactly once. Alternatively, one can allow arbitrary nonnegative distances but the goal is to find a minimum length tour that visits each vertex at least once. Understanding the approximability of the TSP has attracted much attention since Christofides [7] designed a 3/2-approximation algorithm for the problem. Despite great efforts, Christofides algorithm continues to be the current champion, while the best known lower bound, recently obtained by Lampis [13] states that the problem is NP-hard to approximate within a factor 185/184, which improved upon the work of Papadimitriou and Vempala [17]. Very recently Karpinski and Schmied [11] obtained explicit inapproximability bounds for the cases of cubic and subcubic graphs. A key lower bound to study the approximability of the problem is the so-called subtour elimination linear program which has long been known to have an integrality gap of at most 3/2 [26]. A long-standing conjecture (see, e.g., Goemans [9]) states that the integrality gap of the subtour elimination LP is precisely 4/3. There have been several improvements for important special cases of the metric TSP in the last couple of years. Oveis Gharan, Saberi, and Singh [16] designed a (3/2 ɛ)-approximation algorithm for the case of graph metrics, while Mömke and Received by the editors June 16, 2014; accepted for publication (in revised form) March 11, 2015; published electronically May 21, This research was partially supported by Núcleo Milenio Información y Coordinación en Redes ICM/FIC RC An extended abstract of a preliminary version of this work appeared in ESA Departamento de Ingeniería Industrial, Universidad de Chile, Santiago, Chile (correa@uchile.edu, olarre@uchile.edu). Departamento de Ingeniería Matemática and CMM, Universidad de Chile, Santiago, Chile (jsoto@dim.uchile.edu). This author s research was partially supported by FONDECYT grant

2 916 JOSÉ CORREA, OMAR LARRÉ, AND JOSÉ A.SOTO Svensson [14] improved that to 1.461, using a different approach. Mucha [15] then showed that the approximation guarantee of the Mömke and Svensson algorithm is 13/9. Finally, still in the shortest path metric case, Sebö and Vygen [21] found an algorithm with a guarantee of 7/5. These results in particular show that the integrality gap of the subtour LP is below 3/2 in case the metric comes from an unweighted graph. Another notable direction of recent improvements concerns the s-t path version of the TSP on arbitrary metrics, where the natural extension of Christofides heuristic guarantees a solution within a factor of 5/3 of optimum. An, Shmoys, and Kleinberg [2], found a (1 + 5)/2-approximation for this version of the TSP, while Sebö [20] further improved this result obtaining a 8/5-approximation algorithm. This renewed interest in designing algorithms for the TSP in graph metrics has also reached the case when we further restrict ourselves to graph metrics induced by special classes of graphs. Gamarnik, Lewenstein, and Sviridenko [8] showed that in a 3-connected cubic graph on n vertices there is always a TSP tour visiting each vertex at least once of length at most (3/2 5/389)n, improving upon Christofides algorithm for this graph class. A few years later, Aggarwal, Garg, and Gupta [1] improved the result obtaining a bound of (4/3)n while Boyd et al. [5] showed that the (4/3)n bound holds even if only 2-connectivity assumed. Finally, Mömke and Svensson s [14] approach allows us to prove that the (4/3)n bound holds for subcubic 2-connected graphs. Interestingly, the latter bound happens to be tight and thus it may be tempting to conjecture that there are cubic graphs on n vertices for which no TSP tour shorter than (4/3)n exists. The main result in this paper, proved in sections 3 and 4, shows that this is not the case. Namely, we prove that any 2-connected cubic graph on n vertices has a TSP tour of length (4/3 ɛ)n, forɛ =1/61236 > In section 5, we use this result and establish that for cubic graphs, not necessarily 2-connected, there exists a 4/3 ɛ approximation algorithm for the TSP, where ɛ = ɛ/(3 + 3ɛ). Qian et al. [18] showed that the integrality gap of the subtour LP is strictly less than 4/3 for metrics where all the distances are either 1 or 2. Their result, based on the work of Schalekamp, Williamson, and van Zuylen [19], constitutes the first relevant special case of the TSP for which the integrality gap of the subtour LP is strictly less than 4/3. Our result implies that the integrality gap of the subtour LP is also strictly less than 4/3 in connected cubic graphs. From a graph theoretic viewpoint, our result can also be viewed as a step towards resolving Barnette s [4] conjecture, stating that every bipartite, planar, 3-connected, cubic graph is Hamiltonian (a similar conjecture was first formulated by Tait [22], then refuted by Tutte [24], then reformulated by Tutte and refuted by Horton, see, e.g., [6], and finally reformulated by Barnette more than 40 years ago). It is worth mentioning that for Barnette s graphs (i.e., those with the previous properties) on n vertices it is straightforward to construct TSP tours of length at most (4/3)n; however, no better bound was known. Our result improves upon this, and furthermore as we show in section 2, in this class of graphs our bound improves to (4/3 1/18)n <1.28n. In a very recent work, Karp and Ravi [10] proved that for cubic bipartite connected graphs there exists a TSP tour of length at most (9/7)n. This is slightly worse than our (4/3 1/18)n bound but applies to a larger class of graphs, on the other hand the Karp and Ravi result improves upon the bound of (4/3 1/108)n of Larré [12], which applies to cubic bipartite 2-connected graphs Our approach. An Eulerian subgraph cover (or simply a cover) is a collection Γ = {γ 1,...,γ j } of connected multisubgraphs of G, called components, satisfying

3 TSP TOURS IN CUBIC GRAPHS: BEYOND 4/3 917 that (i) every vertex of G is covered by exactly one component, (ii) each component is Eulerian, and (iii) no edge appears more than twice in the same component. Every cover Γ can be transformed into a TSP tour T (Γ) of the entire graph by contracting each component, adding a doubled spanning tree in the contracted graph (which is connected) and then uncontracting the components. Boyd et al [5] defined the contribution of a vertex v in a cover Γ, as z Γ (v) = (l+2) h,whereland h are the number of edges and vertices, respectively, of the component of Γ in which v lies. The extra 2 in the numerator is added for the cost of the double edge used to connect the component to the other in the spanning tree mentioned above, so that v V z Γ(v) equals the number of edges in the final TSP tour T (Γ) plus 2. Let D = {(Γ i,λ i )} k i=1 be a distribution over covers of a graph. This is, each Γ i is a cover of G and each λ i is a positive number so that k i=1 λ i =1. Theaverage contribution of a vertex v with respect to distribution D is defined as z D (v) = k i=1 λ iz Γi (v). As the starting point of our paper, we study short TSP tours on Barnette graphs (bipartite, planar, cubic, and 3-connected). These types of graphs are very special as it is possible to partition their set of faces into three cycle covers, i.e., Eulerian subgraph covers in which every component is a cycle. By a simple counting argument, one of these cycle covers is composed of less than n/6 cycles. By using the transformation described on the previous paragraph, we can obtain a TSP tour of length at most 4n/3. To get shorter TSP tours we describe a simple procedure that performs local operations to reduce the number of cycles in each of the three initial cycle covers. In these local operations we replace the current cycle cover C by the symmetric difference between C and the edges of a given face, provided that the resulting graph is a cycle cover with fewer cycles. By repeating this process on the three initial cycle covers, it is possible to reach a cover with fewer than 5n/36 cycles, which, in turn, implies the existence of a TSP tour of length at most (4/3 1/18)n. This result is described in section 2. The idea of applying local operations to decrease the number of components of an Eulerian subgraph cover can still be applied on more general graph classes. Given a 2-connected cubic graph G, Boyd et al. [5] found a TSP tour T of G with at most 4 3 V (G) 2 edges. Their approach has two phases. In the first phase, they transform G into a simpler cubic 2-connected graph H not containing certain illbehaved structures (called p-rainbows, for p 1). In the second phase, they use a linear programming approach to find a polynomial collection of perfect matchings for H such that a convex combination of them gives every edge a weight of 1/3. Their complements induce a distribution over cycle covers of H. By performing certain local operations on each cover, they get a distribution of Eulerian subgraph covers having average vertex contribution bounded above by 4/3. They use this to find a TSP tour for H with at most 4 3 V (H) 2 edges, which can be easily transformed into a TSP tour of G having the desired guarantee. The local operations used by Boyd et al. [5] consist of iterative alternation of 4-cycles and a special type of alternation of 5-cycles (in which some edges get doubled). Our main result is an improvement on Boyd et al. s technique that allows us to show that every 2-connected cubic graph G admits a TSP tour with at most (4/3 ɛ) V (G) 2 edges. The first difference between the approaches, described in section 3, is that our simplification phase is more aggressive. Specifically, we set up a framework to eliminate large families of structures that we use to get rid of all chorded 6-cycles. This clean-up step can very likely be extended to larger families and may ultimately lead to improved results when combined with an appropriate

4 918 JOSÉ CORREA, OMAR LARRÉ, AND JOSÉ A.SOTO second phase. The second difference, described in section 4, is that we extend the set of local operations of the second phase by allowing the alternation of 6-cycles of a Eulerian subgraph cover. Again, it is likely that one can further exploit this idea to get improved guarantees. Mixing these new ideas appropriately requires significant work but ultimately leads us to find a distribution D of covers of the simplified graph H for which v V (H) z D(v) ( 4 3 ɛ)n 2. From there, we obtain a TSP tour of G with the improved guarantee. Our analysis allows us to set ɛ as 1/61236 > for cubic 2-connected graphs. It is worth noting here that by adding extra hypothesis it is possible to improve this constant. For instance, the case of cubic 2-connected bipartite graphs is interesting. This type of graphs is actually 3-edge colorable; therefore, by taking the complements of the three perfect matchings induced by the coloring, we obtain a cycle cover distribution whose support has only 3 cycle covers making the problem much easier to analyze. In fact, by slightly relaxing the simplification phase we can impose that the resulting graph is still bipartite (but allowing a certain type of chorded hexagons) and thus, the operation consisting on the alternation of 5-cycles will never occur. It is possible to show (see Larré s master s thesis [12]) that this modified algorithm yields a tour of length at most (4/3 1/108)n 2 for any cubic 2-connected bipartite graph. 2. Barnette graphs. Barnette [4] conjectured that every cubic, bipartite, 3- connected planar graph is Hamiltonian. More than 40 years later and despite considerable effort, Barnette s conjecture is still not settled. This motivates the definition of a Barnette graph as a cubic, bipartite, 3-connected planar graph. Even though we are not able to prove or disprove Barnette s conjecture, in this section we show that these graphs admit short tours. Recall that a tour of a graph G is simply a spanning Eulerian subgraph where every edge appears at most twice. The main idea for obtaining short tours is to find a cycle cover C of G having small number of cycles. The tour T (C) obtained by taking the union of the edges of C and a doubled spanning tree of the multigraph obtained by contracting each cycle of C in G has length n +2 C 2, where C is the number of cycles in C. Let G be a Barnette graph on n vertices. By 3-connectedness, G has a unique embedding on the sphere up to isomorphism [25]; therefore, its set of faces is welldetermined. Furthermore, it has a unique planar dual G, which is an Eulerian planar triangulation. As Eulerian planar triangulations are known to be 3-colorable (see, e.g., [23]), Barnette graphs are 3-face colorable. Furthermore, finding such coloring can be done in polynomial time. We can use this to easily get a tour of G of length at most 4n/3 as follows. Denote the vertex, edge, and face sets of G as V = V (G), E = E(G), and F = F (G), respectively, and let c: F {1, 2, 3} be a proper 3-face coloring of G. Let F (i) be the set of faces of color i. Since the graph is cubic, E = 3n/2, and by Euler s formula, F =2+ E V =(n +4)/2. This means that there is a color i such that F (i) (n + 4)/6. Since F (i) is a cycle cover, the tour T (F (i)) obtained as before has length n +2 F (i) 2 (4n 2)/3. We will devise an algorithm that finds a short cycle covers of Barnette graphs by performing certain local operations to reduce the number of cycles of the three cycle covers given by F (1), F (2), and F (3). Recall that an even cycle C 0 is alternating for a cycle cycle cover C of G if the edges of C 0 alternate between edges inside C and edges outside C. If C 0 is an alternating cycle of C, we can define a new cycle cover

5 TSP TOURS IN CUBIC GRAPHS: BEYOND 4/3 919 C C 0 whose edge set is the symmetric difference between the edges of C and those of C 0. The next procedure constructs cycle covers C(i), for each i {1, 2, 3}. Initialize C(i) asf (j) forsomej i; for example, j = i +1(mod3). Aswewillseelater, each face of F (i) is alternating for C(i) at every moment of the procedure. Iteratively, check if there is a face f in F (i) such that C(i) f has fewer cycles than C(i). If so, replace C(i) by the improved cover. Do this step until no improvement is possible to obtain the desired cover. Let C be the cycle cover of fewer cycles among the three covers found. By returning the tour associated to C we obtain Algorithm 1 below. Algorithm 1 To find a tour on a Barnette graph G =(V,E). 1: Find a 3-face coloring of G with colors in {1, 2, 3}. 2: for each i {1, 2, 3} do 3: C(i) F (j), for j =(i +1) (mod3). 4: while there is a face f F (i) such that C(i) f < C(i) do 5: C(i) C(i) f. 6: end while 7: end for 8: Let C be the cycle cover of smallest cardinality among all C(i). 9: Return the tour associated to C. To analyze the algorithm it will be useful to extend the initial face-coloring to an edge-coloring of G, by assigning to each edge e the color in {1, 2, 3} that is not present in both incident faces of e. Denote as E(i) thesetofedgesofcolori. Then E(1), E(2), and E(3) are disjoint perfect matchings and, furthermore, E(i) E(j) are exactly the edges of F (k), for {i, j, k} = {1, 2, 3}. Lemma 1 below implies, in particular, that the algorithm is correct. Lemma 1. In every iteration of Algorithm 1, C(i) is a cycle cover of G containing E(i) and every face f F (i) is alternating for C(i). Proof. We proceed by induction. Observe that C(i) equalsf (j) forj i in the beginning so it contains E(i). Suppose this lemma holds at the beginning of an iteration and let f be a face of F (i), so f has no edges of color i. As C(i) contains E(i), and every vertex v of f has degree 2 in C(i), we conclude that f is alternating for C(i) and thus, C(i) f is a cover containing E(i). Lemma 2. Let C(i) be the cycle cover obtained at the end of the while-loop in Algorithm 1 and let f be a face of length 2k in F (i), thenf intersects at most k+1 2 cycles of C(i) Fig. 1. The graph H in the proof of Lemma 2. Proof. Let C be the collection of cycles in C(i) intersecting f and let H be the subgraph of G whose edge set is the union of the edges of C and those of f. Usingthe +

6 920 JOSÉ CORREA, OMAR LARRÉ, AND JOSÉ A.SOTO planar embedding of G having f as the outer face, we can see that H consists of the outer cycle f and a collection of noncrossing paths on the inside connecting vertices of f as in Figure 1. Label with a plus (+) sign the regions of H bounded by cycles in C(i) andbya minus ( ) sign the rest of the regions except for the outer face, thus C equals the number of + regions. Furthermore, it is easy to see that the cycle cover C(i)Δf has size equal to C(i) minus the number of + regions plus the number of regions. By our algorithm s specification, C(i)Δf C(i), and therefore the number of + regions must be at most that of regions. But since f has length 2k, the total number of regions labeled + and is k+1. Since C is an integer, we get C (k+1)/2. With this lemma we can bound the size of the tour returned. For i {1, 2, 3} and k N define F k (i) as the set of faces of color i and length k, andf k as the total number of faces of length k. Lemma 3. Let C(i) be the cycle cover obtained at the end of the while-loop in Algorithm 1. Then C(i) 1+ k 1 F 2k (i) =1+ F 6 (i) + F 8 (i) +2 F 10 (i) +2 F 12 (i) +. 2 k=2 Proof. Let G be the graph G restricted to the edges of C(i). Let H be the connected Eulerian multigraph obtained by contracting in G all the faces in F (i) to vertices. Observe that the edge set of H is exactly E(i). One by one, uncontract each face f in H. IneachstepweobtainanEuleriangraphwhichmayhavemore connected components than before. We can estimate the number of cycles in C(i) as 1 (the original component in H) plus the increase in the number of connected components on each step of this procedure. Consider the graph H immediately before expanding a face f. Let H f be the connected component of H containing the vertex associated to f. Iff has length 2k, then, after expanding f, H f splits into at most k+1 2 connected components. This follows since after expanding all the cycles of F (i), H f is split into at most that many components by Lemma 2. But then, expanding f increases the number of connected components of H by at most k+1 k 1 2 1= 2. The previous bound is not enough to conclude the analysis. Fortunately, we can find another bound that will be useful. Lemma 4. Let C(i) be the cycle cover obtained at the end of the while-loop in Algorithm 1. If C(i) 1,then k +1 3 C(i) C(i) F 4 + F 2k (i) 2 k=2 = C(i) F 4 + F 4 (i) +2 F 6 (i) +2 F 8 (i) +3 F 10 (i) +. Proof. Since C(i) containe(i), every cycle C C(i) must intersect at least two faces in F(i). In particular, every 4-cycle in C(i) intersects exactly two faces in F(i). Consider now a cycle C C(i) oflengthatleast6. We claim that C must intersect three of more faces of F(i). If this was not the case, then C intersects exactly two faces f 1 and f 2. Then the edges of C must alternate as one edge of f 1, then one edge in E(i) crossing between the faces, then oneedgeinf 2, and one edge in E(i) crossing between the faces. In particular, the length of C must be divisible by 4. Let v 0, v 1,..., v 4k 1 be the vertices of C. By

7 TSP TOURS IN CUBIC GRAPHS: BEYOND 4/3 921 f 2 v 4l+2 v 4l+3 f 1 v 4l v 4l+1 v 4l+4 v 4l+5 Fig. 2. AcycleC intersecting two faces f 1 and f 2. the previous observation we can assume that for every 1 l k, v 4l v 4l+1 is an edge of f 1, v 4l+2 v 4l+3 is an edge of f 2, and the rest of the edges of C are crossing between faces, as depicted in Figure 2. Suppose that there exists a vertex in f 1 not contained in C, then there must be an index l such that v is in the path P from v 4l+1 to v 4l+4 in f 1, that is internally disjoint with C. Sincef 1 is a face of G, we deduce that removing v 4l+1 and v 4l+4 from G disconnects the graph. But this contradicts the 3-connectedness of G. Therefore, C contains all of the vertices in f 1 and, by an analogous argument, all of the vertices of f 2. Since the graph is cubic and connected, the only possibility is that C contains all of the vertices of G. Butthen C(i) = 1 which contradicts the hypothesis of the lemma. Therefore, we have proved the every cycle in C(i) oflengthatleastsix intersects at least 3 faces. Define the set J(i) ={(C, H) C(i) F(i): cycle C intersects face H }. By the previous discussion we have By Lemma 2 we have J(i) 2 C(i) F 4 +3 C(i) \ F 4 =3 C(i) C(i) F 4. J(i) k +1 F 2k (i). 2 k=2 Combining the last two inequalities we get the desired result. Now we have all of the ingredients to bound the size of the cycle cover returned by the algorithm. Lemma 5. Let C be the cycle cover computed by Algorithm 1. Then { n +4 C min F 4 (n +1), + F } Proof. First, we need a bound on a quantity related to previous lemmas. Let α be k 1 F2k. We claim that α 1 2 (n 2 F 4 ). k=2 2 Since F =(n+4)/2, the claim above is equivalent to proving that 2 F 4 +2 F + 4α 3n. Notethat2 F 4 +2 F +4α equals 2 F F F F F F F F F F F F 14 + = 4 F F F F F F 14 +,

8 922 JOSÉ CORREA, OMAR LARRÉ, AND JOSÉ A.SOTO which is upper bounded by k=2 2k F 2k. This quantity is the sum of the length of all the faces in G. As each vertex is in three faces, this quantity is at most 3n, which proves the claim. By Lemma 3 we have C i=1 C(i) 1+ α (n 2 F 4 ) = n +4 6 F 4 6. On the other hand, using that F =(n +4)/2 and Lemma 4, we get 9 C 3 3 C(i) i=1 3 C(i) F 4 + α + F i=1 3 C(i) F (2n +2 F 4 ). Note that each cycle in F 4 avoids one color; therefore, it can only appear in two cycle covers C(i), i.e., 3 i=1 C(i) F 4 2 F 4.Fromherewegetthat i=1 C 1 9 (2 F 4 + n +1 F 4 /2) = n F 4 6. The previous lemma implies the main result of this section. Theorem 1. Let C be the cycle cover computed by Algorithm 1 and let T be the ( tour returned. Then C 5n+14 36, and the length of T is at most 23n = 4 3 ) 1 18 n 11 ( 9. In particular, every Barnette graph admits a tour of length at most 4 3 ) 1 18 n Proof. The expression on the right-hand side of Lemma 5 is maximized when F 4 = n+10 6, for which it attains a value of 5n Therefore, for every value of F 4, this quantity is an upper bound of C. To conclude, we just use that the length of T is n +2( C 1) connected cubic graphs: Simplification phase. We now go back to general 2-connected cubic graphs. Our algorithm starts by reducing the input graph G to a simpler 2-connected cubic graph H which does not contain a cycle of length six with one or more chords as subgraph. In addition our reduction satisfies that if H has a TSP tour of length at most (4/3 ɛ) V (H) 2, then G has a TSP tour of length at most (4/3 ɛ) V (G) 2, where V (H) andv (G) denote the vertex sets of H and G, respectively. We will use the notation χ F {0, 1} E of F E to denote the incidence vector of F (χ F e =1ife F, and 0 otherwise). Reduction 1. Letγ be a 6-cyclehavingtwochordsandlet G[V (γ)] be the subgraph induced by the set of vertices contained in γ Also, let v 1 and v 2 be the two vertices connecting γ to the rest of G. Our reduction replaces G[V (γ)] by a 4-cycle with a chord (shown in Figure 3), identifying v 1 and v 2 with the vertices of degree 2 in the chorded 4-cycle. This procedure in particular removes the p-rainbow structure in Boyd et al. [5]. The second step is to consider 6-cycles having only one chord. Let γ be such a cycle and let G[V (γ)] be the subgraph induced by the set of vertices contained in γ. Consider the four edges e 1, e 2, e 3,ande 4 connecting γ to the rest of G. Letting w i be the endpoint of e i outside γ, we distinguish the following three reductions according to three different cases. Reduction 2. If only two of the w i s are distinct, we proceed as in the previous case (Reduction 1); that is, replacing G[V (γ)] by a chorded 4-cycle.

9 TSP TOURS IN CUBIC GRAPHS: BEYOND 4/3 923 Fig. 3. A 4-cycle with a chord. Reduction 3. If three of the w i s are distinct we replace the 7 vertex structure formed by γ and the w i adjacent to two vertices in γ by a triangle (3-cycle), identifying the degree two vertices in the structure with those in the triangle. Figure 4 shows an example of this reduction in the specific case that γ has a chord connecting symmetrically opposite vertices. Fig. 4. Reduction 3 in the case that γ has a chord connecting symmetrically opposite vertices. Reduction 4. The final case is when all four w i s are distinct. Assume, without loss of generality, that the w i s are indexed in the cyclic order induced by γ. In this case we replace γ by an edge e and we either connect w 1, w 2 to one endpoint of e and w 3, w 4 to the other, or we connect w 1, w 4 to one endpoint and w 2, w 3 to the other. The previous choice can always be made so that in the reduced graph e is not a bridge, as the following lemma shows. Lemma 6. Let γ be a chorded 6-cycle considered in Reduction 4. Then, the edges in G[V (γ)] do not simultaneously contain a cut of G separating the vertex sets {w 1,w 2 } and {w 3,w 4 }, and a second cut separating the vertex sets {w 2,w 3 } and {w 1,w 4 }. Proof. By contradiction, suppose that the edge set of G[V (γ)] contains two cuts of G such that the first cut separates the vertex sets {w 1,w 2 } and {w 3,w 4 },andthe second cut separates the vertex sets {w 2,w 3 } and {w 1,w 4 }. By deleting the vertices and edges of γ, we split G into four connected components, one containing each w i ; thus all paths connecting w 1 and w 3 in G must use the edge e 1, contradicting the 2-connectivity of G. Note that all of the above reduction steps strictly decrease the number of vertices in the graph while keeping the graph cubic and 2-connected, and that each step requires only polynomial time. Thus only a linear number of polynomial time steps are needed to obtain a cubic and 2-connected reduced graph H which does not contain 6-cycles with one or more chords. The following result shows that any TSP tour in the reduced graph H of length at most (4/3 ɛ) V (H) 2 can be turned into a TSP tour in the original graph G of length at most (4/3 ɛ) V (G) 2. Proposition 1. Let T a TSP tour in the reduced graph H of length at most α V (H) 2, with5/4 α 4/3. Then, a TSP tour T can be constructed in the original graph G in polynomial time, such that the length of T is at most α V (G) 2. Proof. We distinguish certain cases, depending on what reduction was performed over the graph. Case of Reduction 1 or 2: in this case the result is a consequence of the following

10 924 JOSÉ CORREA, OMAR LARRÉ, AND JOSÉ A.SOTO lemma. Lemma 7. Let G =(V,E) be a graph and let U V be such that the cut δ(u) has only two elements, say δ(u) ={e 1,e 2 }.Letv, w U be the two end vertices of e 1 and e 2 in U, respectively. Let us suppose that the subgraph G[U] is Hamiltonian and contains a Hamiltonian path connecting v and w. Let H be the graph resulted from replacing the subgraph G[U] by a chorded 4-cycle D and let T be a TSP tour in H. Then, there exists a TSP tour T in G with length T T + U 4. ProofofLemma7. Let χ P be the incidence vector of some Hamiltonian path P connecting v and w, and let χ C be the incidence vector of some Hamiltonian cycle C in G[U]. Let us denote by x the vector χ T (the incidence vector of T ). We are going to extend the TSP tour T to the original graph G depending on the value of x on edges e 1 and e 2. We know that x ({e 1,e 2 }):=x (e 1 )+x (e 2 )mustbeaneven number, since T is a TSP tour. Considering that x takesvaluesover{0, 1, 2}, we have that all of the possible cases for the values of x (e 1 )andx (e 2 ) are as follows. Case (x (e 1 ),x (e 2 )) = (1, 1): in this case there is a path in the chorded 4- cycle D and in T of length 3 connecting v and w. Then, we define the vector x as { χ P (e) if e E(G[U]), x(e) = x (e) if e E(G) \ E(G[U]]). Case (x (e 1 ),x (e 2 )) {(2, 0), (0, 2)}: in this case there is a 4-cycle in D and in T. Then, we define the vector x as { χ C (e) if e E(G[U]), x(e) = x (e) if e E(G) \ E(G[U]). Case (x (e 1 ),x (e 2 )) = (2, 2): in this case we redefine x (e 2 )=0andthenwe define the vector x as the previous case. In any case x is the incidence vector of a TSP tour T in G, oflength T T + U 4. It is straightforward to verify that both Reduction 1 and 2 satisfy the hypothesis of Lemma 7. In the case of Reduction 1, the vertex set of the replaced structure has size U = 6, and in the case of Reduction 2, the vertex set of the replaced structure has size U = 8. Whatever the case, we have that V (G) = V (H) + U 4, and then T T + U 4 α V (H) 2+ U 4 = α V (G) 2 (α 1)( V (G) V(H) ) α V (G) 2, where the first inequality holds by Lemma 7. Case of Reduction 3: in this case the result is a consequence of the following lemma. Lemma 8. Let G =(V,E) a graph and let U V be such that U =7and the cut δ(u) has only three elements, say δ(u) ={e 1,e 2,e 3 }. Let v 1,v 2,v 3 U be the three end vertices of e 1, e 2,ande 3 in U, respectively. Let us suppose that the subgraph G[U] contains a (not necessarily simple) cycle C of length at most 8 which visits every vertex of U, and for every pair of vertices v, w {v 1,v 2,v 3 } there exists

11 TSP TOURS IN CUBIC GRAPHS: BEYOND 4/3 925 a (not necessarily simple) path P (v, w) of length at most 7 which visits every vertex of U. Let H be the graph resulted from replacing the subgraph G[U] by a triangle and let T be a TSP tour in H. Then, there exists a TSP tour T in G with length T T +5. Proof of Lemma 8. Let χ P (v, w) be the incidence vector of path P (v, W) and χ C be the incidence vector of C. Let us denote by x the vector χ T (the incidence vector of T ). We are going to extend the TSP tour T to the original graph G depending on the value of x on edges e 1, e 2,ande 3. We know that x ({e 1,e 2,e 3 }):= x (e 1 )+x (e 2 )+x (e 3 ) must be an even number, since T is a TSP tour. Considering that x takes values over {0, 1, 2}, we have that all of the possible cases for the values of x (e 1 ), x (e 2 ), and x (e 3 ) are as follows. Case (x (e 1 ),x (e 2 ),x (e 3 )) = (2, 0, 0) (or another possible permutation): in this case we define the vector x as { χ C (e) if e E(G[U]), x(e) = x (e) if e E(G) \ E(G[U]). Case (x (e 1 ),x (e 2 ),x (e 3 )) {(2, 2, 0), (2, 2, 2)} (or another possible permutation): first we redefine x (e 2 )=x (e 3 ) = 0 and then we define the vector x as the previous case. Case (x (e 1 ),x (e 2 ),x (e 3 )) = (1, 1, 0) (or another possible permutation): in this case we define the vector x as { χ P (v1,v2) (e) if e E(G[U]), x(e) = x (e) if e E(G) \ E(G[U]). Case (x (e 1 ),x (e 2 ),x (e 3 )) = (1, 1, 2) (or another possible permutation): first we redefine x (e 3 ) = 0 and then we define the vector x as the previous case. In any case x is the incidence vector of a TSP tour T in G. Note that x was constructed from x, by adding edges from the original graph and removing edges from the triangle. In the first and second case we added at most eight edges and deleted at least 3. In the third and fourth cases we added at most seven edges and deleted at least 2. In any case, we have that T T +5. To see that all possible structures that are considered in Reduction 3 satisfy the hypothesis of Lemma 8, we need only check by inspection that there exist a cycle C of length at most 8 and paths of length at most 7 connecting every pair of vertices, which visit every vertex of the structures of Figure 5. Then, applying Reduction 3, we can define a TSP tour T in the original graph with length T T +5 α V (H) 2+5 = α V (G) 2+5+α( V (H) V(G) ) = α V (G) 2+(5 α4) α V (G) 2, where the first inequality holds by Lemma 8 and the last one since 5/4 α. Case of Reduction 4: it is easy to construct a TSP tour T in the original graph G with length T T + 4. The detailed case analysis is omitted since it is very

12 926 JOSÉ CORREA, OMAR LARRÉ, AND JOSÉ A.SOTO Fig. 5. All possible structures that are considered in Reduction 3. The three figures on the top consider the situations in which γ has a chord dividing the cycle into two 4-cycles and the extra vertex connects two vertices in γ at distance 1, 2, and3. The next four cases correspond to the situations in which γ has a chord dividing the cycle into a 3-cycle and a 5-cycle and the extra vertex connects two vertices in γ at distance 1 (middle right), at distance 2 (left and right) and at distance 3 (middle left). Symmetric cases are displayed only once. similar to the previous ones. Then T T +4 α V (H) 2+4 = α V (G) 2+4+α( V (H) V(G) ) = α V (G) 2+(4 α4) α V (G) 2, where the last inequality holds since α 1. Considering this latter case, we finish the proof of Proposition 1. Finally, note that as mentioned above only a linear number of reduction steps are needed, and each step requires only polynomial time, not only to find the desired structure, but also to recover the TSP tour in the original graph. Thus this graph simplification phase runs in polynomial time connected cubic graphs: Eulerian subgraph cover phase. We say that a matching M is 3-cut perfect if M is a perfect matching intersecting every 3-cut in exactly one edge. Boyd et al. [5] have shown the following lemma. Lemma 9 (see [5]). Let G =(V,E) be a 2-connected cubic graph. Then, the vector 1 3 χe can be expressed as a convex combination of incidence vectors of 3-cut perfect matchings of G. Thisis,thereare3-cut perfect matchings {M i } k i=1 and positive real numbers λ 1,λ 2,...,λ k such that (4.1) and (4.2) k λ i =1 i=1 1 3 χe = k λ i χ Mi. i=1

13 TSP TOURS IN CUBIC GRAPHS: BEYOND 4/3 927 Fig. 6. Operation (U1). Furthermore, Barahona [3] provides an algorithm to find a convex combination of 1 3 χe having k 7n/2 1 in O(n 6 ) time. Consider a graph G that is cubic, 2-connected, and reduced. That is, no 6-cycle in G has chords. We also assume that n 10 as every cubic 2-connected graph on less than ten vertices is Hamiltonian. Let {M i } k i=1 and {λ i} k i=1 be the 3-cut matchings and coefficients guaranteed by Lemma 9. Let {C i } k i=1 be the family of cycle covers associated to the matchings {M i } k i=1.thisis,c i is the collection of cycles induced by E \ M i. Since each matching M i is 3-cut perfect, the corresponding cycle cover C i does not contain 3-cycles. Furthermore, every 5-cycle in C i is induced (i.e., it has no chord in G). In what follows we define three local operations, (U1), (U2), and (U3), that will be applied iteratively to the current family of covers. Each operation is aimed to reduce the contribution of each component of the family. We stress here that operations (U2) and (U3) are exactly those used by Boyd et al. [5], but for the reader s convenience we explain them here. We start with operation (U1). (U1) Consider a cycle cover C of the current family. If C 1, C 2,andC 3 are three disjoint cycles of C that intersect a fixed 6-cycle C of G (which, because of the graph simplification phase, has no chords), then we merge them into the simple cycle obtained by taking their symmetric difference with C. Thisis, the new cycle in V (C 1 ) V (C 2 ) V (C 3 ) having edge set (E(C 1 ) E(C 2 ) E(C 3 ))ΔE(C). An example of (U1) is depicted in Figure 6. We apply (U1) as many times as possible to get a new cycle cover {Ci U1 } k i=1. Then we apply the next operation. (U2) Consider a cycle cover C of the current family. If C 1 and C 2 are two disjoint cycles of C that intersect a fixed 4-cycle C of G, then we merge them into a simple cycle obtained by taking their symmetric difference with C. Thisis, the new cycle in V (C 1 ) V (C 2 ) having edge set (E(C 1 ) E(C 2 ))ΔE(C). We apply operation (U2) as many times as possible to obtain a new cycle cover {Ci U2 } k i=1 of G. The next operation we define may transform a cycle cover C of the current family into a Eulerian subgraph cover Γ, having components that are not necessarily cycles. (U3) Let Γ be an Eulerian subgraph cover of the current family. If γ 1 and γ 2 are two components of Γ, each one having at least five vertices, whose vertex set intersect a fixed 5-cycle C of G, then combine them into a single component, by adding at most one extra edge. To explain how we combine the components in operation (U3) we need the following two lemmas. Lemma 10 (see [5]). Let H 1 and H 2 be two connected Eulerian multisubgraphs of a cubic graph G having at least two vertices in common and let H 3 be the sum of H 1 and H 2, i.e., the union of their vertices and the sum of their edges (allowing multiple parallel edges). Then we can remove (at least) two edges from H 3 such that it stays

14 928 JOSÉ CORREA, OMAR LARRÉ, AND JOSÉ A.SOTO Fig. 7. Sketch of operation (U3). connected and Eulerian. Proof. Letu and v be in both H 1 and H 2.TheedgesetofH 3 can be partitioned into edge-disjoint (u, v)-walks P 1, P 2, P 3,andP 4. Since u has degree 3 in G, there must be two parallel edges incident to u that are on different paths, say e 1 P 1 and e 2 P 2. If we remove e 1 and e 2, then the graph stays Eulerian. Moreover, it stays connected since u and v are still connected by P 3 and P 4 and every vertex of P 1 and P 2 is still connected to one of u and v. Lemma 11 (similar to an observation in [5]). If v belongs to a component γ of any of the covers Γ considered by the algorithm, then at least two of its three neighbors are in the same component. Proof. The above lemma holds trivially when γ is a cycle. In particular, the lemma holds before the application of operation (U3). As the vertex set of a component created by operation (U3) is the union of the vertex set of two previous components, the above lemma also holds after operation (U3). Observe that if γ is a component of a cover in the current family, and C is an arbitrary cycle of G containing a vertex of γ, then, by the cubicity of G and Lemma 11, C and γ must share at least two vertices. In particular, if γ 1 and γ 2 are the two components intersecting a 5-cycleC considered by operation (U3), then one of them, say γ 1, must contain exactly two vertices of C and the other one must contain the other three vertices (note that they cannot each share two vertices, since then avertexofc would not be included in the cover). To perform (U3) we first merge γ 1 and C using Lemma 10 removing two edges, and then we merge the resulting component with γ 2, again removing two edges. Altogether, we added the five edges of C and removed four edges. Finally, we remove two edges from each group of triple or quadruple edges that may remain, so that each edge appears at most twice in each component. Figure 7 shows an example of (U3). Remark 1. Operation (U3) generates components having at least ten vertices. Therefore, any component having nine or fewer vertices must be a cycle. Furthermore, all the cycles generated by (U1) or (U2) contain at least ten vertices (this follows from the fact that G is reduced, and so operation (U2) always involves combining two cycles of length at least 5). From here we observe that any component having nine or fewer vertices must be in the original cycle cover {C i } k i=1. We say that a 4-cycle C with a chord is isolated if the two edges incident to it are not incident to another chorded 4-cycle. The following is the main result of this section. Before proving it we show it implies the main result of this paper. Proposition 2 (main proposition). Let {Γ i } k i=1 be the family of Eulerian subgraph covers at the end of the algorithm (that is, after applying all operations), and let z(v) = z D (v) be the average contribution of vertex v for the distribution D = {(Γ i,λ i )} k i=1. Furthermore, let γ i be the component containing v in Γ i and Γ(v) ={γ i } k i=1. We have the following. (P1) If v is in an isolated chorded 4-cycle, then z(v) 4/3. (P2) If v is in a nonisolated chorded 4-cycle of G, thenz(v) 13/10. (P3) Else, if there is an induced 4-cycle γ Γ(v), thenz(v) 4/3 1/60.

15 TSP TOURS IN CUBIC GRAPHS: BEYOND 4/3 929 (P4) Else, if there is an induced 5-cycle γ Γ(v), thenz(v) 4/3 1/60. (P5) Else, if there is an induced 6-cycle γ Γ(v), then we have both z(v) 4/3 and w V (γ) z(w) 6 (4/3 1/729). (P6) In any other case z(v) 13/10. Theorem 2. Every 2-connected cubic graph G =(V,E) admits a TSP tour of length at most (4/3 ɛ) V 2, whereɛ =1/ This tour can be computed in polynomial time. ProofofTheorem2. From section 3, we can assume that G is also reduced and so the main proposition holds. Let B be the union of the vertex sets of all isolated chorded 4-cycles of G. We say a vertex is bad if it is in B, andgood otherwise. We claim that the proportion of bad vertices in G is bounded above by 6/7. To see this, construct the auxiliary graph G from G by replacing every isolated chorded 4-cycle with an edge between its two neighboring vertices. Since G is cubic, it contains exactly 2 E(G ) /3 vertices, which are good in G. Hence, for every bad vertex there are at least (1/4) (2/3) = 1/6 good ones, proving the claim. The Main proposition guarantees that every bad vertex v contributes a quantity z(v) 4/3. Now we show that the average contribution of all the good vertices is at most (4/3 δ) forsomeδ to be determined. To do this, define H = {γ i Γ i : V (γ) =6} as the collection of all 6-cycles appearing in some cover of the final family, and let H = γ H V (γ) be the vertices included in some 6-cycle of H. It is easy to check that B and H are disjoint. Furthermore, the Main proposition guarantees that if v V \(B H), then z(v) (4/3 1/60). So we focus on bounding the contribution of the vertices in H. For every v H, letf(v) be the number of distinct 6-cycles in H containing v. Since G is cubic, there is an absolute constant K, such that f(v) K. BytheMain proposition, z(v) 4/3 forv H and for every γ H, v V (γ) z(v) 6 (4/3 ɛ ), where ɛ =1/729. Putting this all together we have that K [ ( )] 4 z(v) 3 ɛ = H ɛ + K ( z(v) 4 ) K 3 v H v H 6 H ɛ + ( f(v) z(v) 4 ) =6 H ɛ + ( z(v) 4 ) 3 3 v H γ H 6 H ɛ γ H 6ɛ =0. v V (γ) 1 It follows that H v H z(v) (4/3 ɛ /K). Since ɛ /K 1/60, we get z(v) z(v)+ z(v)+ z(v) v V v B v H v V \(B H) 4 ( ) ( ) 4 3 B + 3 ɛ 4 ( V B ) = V K 3 ɛ. 7K We conclude that there is an index i such that v V z i(v) V (4/3 ɛ /(7K)). By adding a double spanning tree of G/E(Γ i ) we transform Γ i into a TSP tour T of length V (4/3 ɛ /(7K)) 2. Noting that K 12 and ɛ =1/729 we obtain the desired bound. 1 Clearly, all operations can be done in polynomial time. 1 Consider two edges, e 1 and e 2, adjacent to v. Since there is no chorded 6-cycle, if e 1 and e 2 are contained in a 4-cycle, then v must be contained in at most one 6-cycle. Otherwise, there are at most four 6-cycles which may contain e 1 and e 2. Because there are three possible pairs of edges, we have K = 12.

16 930 JOSÉ CORREA, OMAR LARRÉ, AND JOSÉ A.SOTO 4.1. Proof of the main proposition. We start by a lemma, whose proof is the same as that of [5, Observation 1]. Lemma 12 ([5]). For each vertex v V,andeachi {1,...,k}, the contribution z i (v) :=z Γi (v) is (a) at most h+2 h,whereh =min{t, 10} and v is on a t-cycle belonging to one of the cycle covers C i, Ci U1,andCi U2. (b) at most if operation (U3) modified the component containing v. We will also use the following notation in our proof. For any subset J of indices in [k] :={1,...,k}, define λ(j) = i J λ i. The proofs of parts (P1) through (P4) are similar to the arguments used by Boyd et al. [5] to show that z(v) 4/3 whenv is a 4-cycle or 5-cycle. By using the fact that G is reduced (i.e., it contains no chorded 6-cycles) we obtain a better guarantee in (P2), (P3) and (P4). To prove part (P5) we heavily use the fact that operation (U1) is applied to the initial cycle cover (recall that this operation was not used in [5]) Proof of part (P1) of the main proposition. Let v be in some isolated chorded 4-cycle C with V (C) ={a, b, u 0,u 1 } as in Figure 8. Fig. 8. A chorded 4-cycle. For every index i, letc i be the cycle containing v in the initial cycle cover C i, and let C(v) ={C i } k i=1. Consider a cycle C C(v), and recall that C cannot be a triangle. If C does not contain the edge u 0 u 1,thenC = C. Consider now the case in which C contains u 0 u 1. Then we must also have ab E(C )andv 0 v 1 E(C ). Since the graph is reduced, v 1 u 1 / E as otherwise u 1 u 0 a b v 0 v 1 would induce a chorded 6-cycle. Hence, the cycle C cannot be of length 6. It also cannot be of length 7 since then there would be a 3-cut with three matching edges. Therefore, it must be of length at least 8. Using that {i : u λ 1u 2 M i} i = 1 3 and applying Lemma 12, we conclude that z(v) (1/3 6/4+2/3 10/8) = 4/ Proof of part (P2) of the main proposition. Let v be in some nonisolated chorded 4-cycle C with V (C) ={a, b, u 0,u 1 } as in Figure 8 and recall that v 1 u 1 E. Without loss of generality, we can assume that u 1 is in a different chorded 4-cycle D. Furthermore, assume that v 1 is not connected by an edge to D, asthis would imply the existence of a bridge in G. Consider, as in the proof of part (P1) a cycle C C(v). If C does not contain the edge u 0 u 1,thenC = C. If on the other hand the edge v 0 v 1 is in C,thenC must contain all the vertices of both C and D. Itmustalsocontainv 1 and one of its neighbors outside C D. In particular, C has at least ten vertices. By Lemma 12, we have that z(v) (1/3 6/4+2/3 12/10) = 13/ Proof of part (P3) of the main proposition. Let γ Γ(v) bean induced 4-cycle containing v. ByRemark1,γ is in some initial cycle cover C i.since the cycle γ has no chord, then the four edges incident to it (i.e., those sharing one vertex with γ) belongtomatchingm i. This observation holds not only for γ but for

17 TSP TOURS IN CUBIC GRAPHS: BEYOND 4/3 931 Fig cycle γ intersecting an 8-cycle. any cycle C in some initial cycle cover C i,sowehavethefollowingremark. Remark 2. Let P be a path not sharing edges with a cycle C belonging to some initial cycle cover C i. If P connects any two vertices of C,thenP has length at least 3. Furthermore, as the graph is reduced, γ does not share exactly one edge with any other 4-cycle (as this would induce a 6-cycle with a chord). In other words we have the following property. Remark 3. Let P be a path not sharing edges with γ. IfP connects any pair of consecutive edges of γ, thenp has length at least 4. Define the sets X p = {i: C M i = p}, forp =0, 1, 2 and note that X 0 X 1 X 2 = [k]. Define also x p = λ(x p ), for p =0, 1, 2. By (4.1), we have x 0 + x 1 + x 2 = 1. Also, by applying (4.2) to the set of four edges incident to γ we obtain 4x 0 +2x 1 =4/3, which implies that x 0 =1/3 x 1 /2. Finally, by applying (4.2) to the four edges inside γ, weobtainx 1 +2x 2 =4/3, which implies that x 2 =2/3 x 1 /2. For every i X 0, the cycle containing v in C i is equal to γ. By Lemma 12 we obtain z i (v) 6/4 =3/2. Using Remark 3 we deduce that for every i X 1, the cycle containing v in C i has length at least 7; therefore, by Lemma 12, we have z i (v) 9/7. Consider now an index i X 2. Suppose that γ intersects two different cycles of C i. As each of them has length at least 5 and they both share one edge with a 4-cycle of G we conclude that both cycles are modified by operation (U1) or (U2). Remark 1 implies that v is in a cycle of length at least 10 in Ci U2. Using Lemma 12 we have z i (v) 12/10 = 6/5. The only remaining case is if γ is intersected by a single cycle C of C i. Then, by Remark 3, C has length at least 8. This cycle has length exactly 8 if and only if γ belongs to the structure depicted in Figure 9. Assume for now that no 8-cycle of an initial cover contains the four vertices of γ. Then, the cycle C in our previous discussion must be of length at least 9, and by Lemma 12, z i (v) max{11/9, 6/5} =11/9. Putting this all together, we obtain z(v) x 0 3/2+x 1 9/7+x 2 11/9 =(1/3 x 1 /2)3/2+x 1 9/7+(2/3 x 1 /2)11/9 =71/54 + x 1 (9/7 3/4 11/18) 71/54 = 4/3 1/54. Now consider the case in which there is an 8-cycle C j of an initial cover C j containing V (γ). Then v belongs to the structure depicted in Figure 9, where e 1 e 3, e 2 e 4,ande 1,e 2,e 3,e 4 are in some matching M j. As we assumed that V (G) 10, we cannot simultaneously have e 1 = e 4 and e 2 = e 3.Letf and g be the leftmost and